Recurrence Relations and Inequalities Stefan Gerhold RISC - - PDF document

Recurrence Relations and Inequalities

Stefan Gerhold RISC (Research Institute for Symbolic Computation) Linz/Hagenberg, Austria

Proving Inequalities by Computer Algebra

(with M. Kauers, ISSAC 2005)

• We want to prove an > 0, n ≥ 0
• an polynomially recursive (not necessarily holonomic)

an+s = P(an, . . . , an+s−1), n ≥ 0. Example: 22n

• induction step

an > 0, . . . , an+r−1 > 0 = ⇒ an+r > 0

• Sufficient:

X0 > 0, . . . , Xr−1 > 0 = ⇒ Xr > 0 (∗) for all real numbers X0, . . . , Xr that satisfy polyno- mial equations arising from the recurrence of an.

• Increase r, if formla (∗) does not hold;
• r encode known inequalities/identities as additional

inequalities/equations for the Xk.

Proving Inequalities by Computer Algebra

• an may involve undetermined parameters
• Works on many examples (Cauchy-Schwarz, Bernoulli,

Tur´ an, . . . )

• But no a priori termination criterion known
• Example gallery (omitting some constraints):

(x + 1)n ≥ 1 + nx, n ≥ 0, x ≥ −2 (!) Pn(x)2 − Pn−1(x)Pn+1(x) ≥ 0 n

• k=1

xkyk 2 ≤

n

• k=1

x2

k n

• k=1

y2

k n

• k=1

(1 − ak) > 1 −

n

• k=1

ak (n − 1) n

• k=1

ak 2 ≥ 2n

n

• k=1

ak

k−1

• i=1

ai

• n +
• (n − 1) +
• · · · +
• 2 +

√ 1 < √n + 1

Linear Recurrences with Constant Coef- ficients

• Is there a subclass over which positivity is decidable?
• C-finite sequences are a natural candidate:

c0an + c1an+1 + · · · + cdan+d = 0, n ≥ 0.

• Question: an > 0, n large?
• power sum representation

an =

d

• k=1

bkαn

k,

αk roots of characteristic polynomial (assumed to be simple here).

• dominating roots := roots of largest modulus
• Usually the sign is trivial:

3n + (−2)n > 0 (−3)n + (−2)n ≶ 0

Sequences with no Positive Dominating Root

• What is the sign of (2 + i)n + (2 − i)n?
• (Side remark: Lower estimate

5n/2/nC < |(2 + i)n + (2 − i)n| Schinzel 1967, based on Baker’s theorem about linear forms in logarithms)

• Conjecture: infinitely many sign changes, if no dom-

inating root is positive.

• 1 dominating root: trivial (root has to be real nega-

tive)

• 2 dominating roots: Burke, Webb 1981

Example: (2 + i)n + (2 − i)n ≶ 0

• 3 or 4 dominating roots: SG 2005 (By Diophantine

geometry) Example: (− √ 5)n + (2 + i)n + (2 − i)n ≶ 0

Sequences with no Positive Dominating Root: Two Pairs of Conjugated Roots

• an = w1 sin(2πnθ1 + ϕ1) + w2 sin(2πnθ2 + ϕ2)
• Example: (θ1, θ2) = ( 7

10, 1 5)

• The set {(7n/10, n/5) mod 1 : n ∈ ◆}
• Every square with side length 1/2, parallel to the

axes, contains a point (Minkowsi’s theorem from Dio- phantine geometry)

• Hence (sin(2πnθ1 +ϕ1), sin(2πnθ2 +ϕ2)) assumes all

four sign combinations (+1, +1), (+1, −1), (−1, +1), (−1, −1).

• Hence an oscillates for all w1, w2 (not both zero).

Sequences with no Positive Dominating Root: General Case

• Theorem (Bell, SG 2005): Let an be a C-finite se-

quence, not identically zero, with no positive dom- inating root. Then the sets {n : an > 0} and {n : an < 0} have positive density.

• Theorem (Kronecker, Weyl): The sequence (nθ1, . . . , nθm)

is uniformly distributed modulo 1, if the numbers 1, θ1, . . . , θm are linearly independent over ◗.

• Idea: Study the function of t resulting from an upon

replacing each nθk by a real tk.

• The other “extreme case”: all θk are rational. Let q

be a common denominator.

q−1

• j=0

aj =

q−1

• j=0

m

• i=1

wi sin(2πjθi + ϕi) =

m

• i=1

wi

q−1

• j=0

(cos ϕi · sin 2πjθi + sin ϕi · cos 2πjθi) = 0.

Which Numbers Occur as Density of the Positivity Set?

• No positive dominating root: Numerical experiments

usually yield approximations ≈ 1/2.

• Theorem: For every κ ∈ ]0, 1[, there is a C-finite

sequence an with no positive dominating root and density({n ∈ ◆ : an > 0}) = κ.

• With positive dominating root: all numbers from

[0, 1] occur.

• Theorem: r rational, 0 ≤ κ, r ≤ 1, κ+r ≤ 1. Then

there is a C-finite sequence an with density({n ∈ ◆ : an > 0}) = κ, density({n ∈ ◆ : an = 0}) = r.

• r must be rational by the Skolem-Mahler-Lech the-
• rem (zero set is periodic).

Conclusion

• Inequalities are more difficult than identities
• Still, many examples can be done by the method we

presented

• But even positivity of C-finite sequences is not known

to be decidable

• Diophantine methods reveal oscillating behaviour
• Problem: Do we have

1 + sin(2πθn) + (−1/2)n > 0, where θ = √ 2? (Remark: for almost all θ, this sequence is eventually positive.)